Question

How do you prove the identity `(sinx-cosx)/cos^2x=(tan^2x-1)/(sinx+cosx)`?

Answer 1

See explanation...

Explanation:

The difference of squares identity can be written:

`a^2-b^2 = (a-b)(a+b)`

Use this with `a = sin x` and `b = cos x` to find:

`(sin x - cos x)/(cos^2 x) = ((sin x - cos x)(sin x + cos x))/((cos^2 x)(sin x + cos x)`

`color(white)((sin x - cos x)/(cos^2 x)) = (sin^2 x - cos^2 x)/((cos^2 x)(sin x + cos x)`

`color(white)((sin x - cos x)/(cos^2 x)) = ((sin^2 x)/(cos^2 x) - (cos^2 x)/(cos^2 x))/(sin x + cos x)`

`color(white)((sin x - cos x)/(cos^2 x)) = (tan^2 x - 1)/(sin x + cos x)`